CRACK IBPS Clerk Prelims – Quadratic Equation Day 25

October 28, 2017

CRACK IBPS Clerk Prelims – Quadratic Equation Day 25

Dear Bankersdaily Zealot ,

IBPS Clerk preliminary examination will be held in the month of December this year and you know the preparations are to be in sky level to reach the perfect destination of achieving the feat. Daily Quizzes enrich your preparations to the next level and you could also learn the mistakes at a daily pace.

This clerk planner is based on the previous year patterns and this daily planner which is going to be conducted for a period of 50 days will provide you the pace to attend the exam in the preliminary stage. This daily tests will be conducted in a quiz format as like the IBPS PO planner which we have conducted earlier.

Perpetual preparations will be the key to crack the exam. So ,don’t forget to check the daily quizzes on various topics in different sections in the planner. Do note the time , as we have allotted different timings for each section.

We hope these tests will entice you to practice more. If you face any issues in solving these problems, we are always here to help you. Please comment your doubts in the comments section , So that you get your doubts clarified instantly.

Do follow the tests regularly and we hope this will help you to crack the exams. At the end , you will also be able to attend free mock tests which will be in the same pattern as that of the exams.

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IBPS Clerk Study Planner

Quadratic Equation day 25

Time : 10 Minutes

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Question 1 of 10

1. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. 3x2 – 29x + 56 = 0

II. 3y2 – 5y – 8 = 0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(b)

3x2 – 29x + 56 = 0

or 3x2 – 21x – 8x + 56 = 0

or 3x(x – 7) – 8(x – 7) = 0

or (3x – 8) (x – 7) = 0

x = , 7

II. 3y2 – 5y – 8 = 0

or 3y2 + 3y – 8y – 8 = 0

or 3y(y + 1) – 8(y + 1) = 0

or (3y – 8) (y + 1) = 0

or (3y – 8) (y + 1) = 0

y = -1,

x≥ y

Incorrect

(b)

3x2 – 29x + 56 = 0

or 3x2 – 21x – 8x + 56 = 0

or 3x(x – 7) – 8(x – 7) = 0

or (3x – 8) (x – 7) = 0

x = , 7

II. 3y2 – 5y – 8 = 0

or 3y2 + 3y – 8y – 8 = 0

or 3y(y + 1) – 8(y + 1) = 0

or (3y – 8) (y + 1) = 0

or (3y – 8) (y + 1) = 0

y = -1,

x≥ y

Question 2 of 10

2. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. 5x2 + 26x – 24 = 0

II. 5y2 – 34y + 24 = 0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(d)

5x2 + 26x – 24 = 0

or 5x2 + 30x – 4x – 24 = 0

or 5x(x + 6) – 4(x + 6) = 0

or (5x – 4) (x + 6) = 0

x = 4/5, 6

II. 5y2 – 30y – 4y + 24 = 0

or 5y(y – 6) – 4(y – 6) = 0

or (5y – 4) (y – 6) = 0 4

y =4/5 6

x≤y

Incorrect

(d)

5x2 + 26x – 24 = 0

or 5x2 + 30x – 4x – 24 = 0

or 5x(x + 6) – 4(x + 6) = 0

or (5x – 4) (x + 6) = 0

x = 4/5, 6

II. 5y2 – 30y – 4y + 24 = 0

or 5y(y – 6) – 4(y – 6) = 0

or (5y – 4) (y – 6) = 0 4

y =4/5 6

x≤y

Question 3 of 10

3. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. x2– 7x = 0

II. 2y2 + 5y + 3 = 0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(a)

x2 – 7x = 0

or x (x – 7) = 0

x = 0, 7

II. 2y2 + 5y + 3 = 0

or 2y2 + 2y + 3y + 3 = 0

or 2y(y + 1) + 3(y + 1) = 0

or (2y + 3) (y + 1) = 0

y = -1, -3/2

x > y

Incorrect

(a)

x2 – 7x = 0

or x (x – 7) = 0

x = 0, 7

II. 2y2 + 5y + 3 = 0

or 2y2 + 2y + 3y + 3 = 0

or 2y(y + 1) + 3(y + 1) = 0

or (2y + 3) (y + 1) = 0

y = -1, -3/2

x > y

Question 4 of 10

4. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. 7x – 4y = 40

II. 8x + 8y = 8

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(a)

7x – 4y = 40 …(i)

and 8x + 8y = 8

or x+y= 1 …(ii)

Solving (i) and (ii), we have

x = 4, y = -3

x > y

Incorrect

(a)

7x – 4y = 40 …(i)

and 8x + 8y = 8

or x+y= 1 …(ii)

Solving (i) and (ii), we have

x = 4, y = -3

x > y

Question 5 of 10

5. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. 15x2 – 41x + 14 = 0

II. 2y2 – 13y + 20 = 0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(c)

15x2 – 4x + 14 = 0

or 15x2 – 6x – 35x + 14 = 0

or 3x(5x – 2) – 7(5x – 2) = 0

or (3x – 7)(5x – 2) = 0

x = 7/3, 2/5

II. 2y2 – 13y + 20 = 0

or 2y2 – 8y – 5y + 20 = 0

or 2y(y – 4) – 5(y – 4) = 0

or (2y – 5) (y – 4) = 0

y = 4, 5/2

x < y

Incorrect

(c)

15x2 – 4x + 14 = 0

or 15x2 – 6x – 35x + 14 = 0

or 3x(5x – 2) – 7(5x – 2) = 0

or (3x – 7)(5x – 2) = 0

x = 7/3, 2/5

II. 2y2 – 13y + 20 = 0

or 2y2 – 8y – 5y + 20 = 0

or 2y(y – 4) – 5(y – 4) = 0

or (2y – 5) (y – 4) = 0

y = 4, 5/2

x < y

Question 6 of 10

6. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. x2-8√3x+45=0

II. y2-√2y-24=0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(e)

I.x2 -8√3x + 45= 0

or x2 – 5√3x + 3√3 (x – 5√3) = 0

or, (x + 3√3) (x – 5√3) = 0

X = 3√3, 5√3

II. y2 – √2y – 24 = 0

Or y2 – 4√2y + 3 √2y – 24 = 0

Or (y-4√2y) (y + 2√2)

y = -3 √2, 4√2

Hence relation cannot be established between x and y.

Incorrect

(e)

I.x2 -8√3x + 45= 0

or x2 – 5√3x + 3√3 (x – 5√3) = 0

or, (x + 3√3) (x – 5√3) = 0

X = 3√3, 5√3

II. y2 – √2y – 24 = 0

Or y2 – 4√2y + 3 √2y – 24 = 0

Or (y-4√2y) (y + 2√2)

y = -3 √2, 4√2

Hence relation cannot be established between x and y.

Question 7 of 10

7. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. x-7√2x+24=0

II. y-5√2y+12=0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

x – 7 √2x + 24 = 0

Or x – 4√2x – 3 √2x + 24 = 0

Or √x (√x – 4√2) – 3√2 (√x – 4√2) = 0

Or (√x – 3√2) (√x – 4√2) = 0

Now, if √x -3√2 = 0

then √x = 3√2

x = 9 × 2 = 18

If √x – 4√2 = 0

then √x = 4√2

x = 16 × 2 = 32

II. y – 5√2y + 12 = 0

y -3√2y – 2√2y + 12 = 0

Or √y (√y – 3√2) -2√2y + 12 = 0

Or (√y – 2√2) – (√y -3√2) = 0

If (√y -2√2) = 0

Then √y = 2√2

y = 4 2 = 18

If √y -3√2 = 0

Then, √y -√2

y = 9 × 2 = 18

x ≥ y

Incorrect

x – 7 √2x + 24 = 0

Or x – 4√2x – 3 √2x + 24 = 0

Or √x (√x – 4√2) – 3√2 (√x – 4√2) = 0

Or (√x – 3√2) (√x – 4√2) = 0

Now, if √x -3√2 = 0

then √x = 3√2

x = 9 × 2 = 18

If √x – 4√2 = 0

then √x = 4√2

x = 16 × 2 = 32

II. y – 5√2y + 12 = 0

y -3√2y – 2√2y + 12 = 0

Or √y (√y – 3√2) -2√2y + 12 = 0

Or (√y – 2√2) – (√y -3√2) = 0

If (√y -2√2) = 0

Then √y = 2√2

y = 4 2 = 18

If √y -3√2 = 0

Then, √y -√2

y = 9 × 2 = 18

x ≥ y

Question 8 of 10

8. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. 12x2 – 17x + 6 = 0

II. 20y2 – 31y + 12 = 0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(d)

12x2 – 17x + 6 = 0

or 12x2 – 9x – 8x + 6 = 0

or 3x(4x – 3) – 2(4x – 3) = 0

or (3x – 2) (4x – 3) = 0

If 3x – 2 = 0

then 3x = 2

x= 2/3

If 4x – 3 = 0

then x = 3/4

II. 20y2 – 31y + 12 = 0

or 20y2 – 16y – 15y + 12 = 0

or 4y (5y – 4) – 3 (5y – 4) = 0

or (4y – 3) (5y – 4) = 0

y=3/4,4/5

Hence x ≤ y

Incorrect

(d)

12x2 – 17x + 6 = 0

or 12x2 – 9x – 8x + 6 = 0

or 3x(4x – 3) – 2(4x – 3) = 0

or (3x – 2) (4x – 3) = 0

If 3x – 2 = 0

then 3x = 2

x= 2/3

If 4x – 3 = 0

then x = 3/4

II. 20y2 – 31y + 12 = 0

or 20y2 – 16y – 15y + 12 = 0

or 4y (5y – 4) – 3 (5y – 4) = 0

or (4y – 3) (5y – 4) = 0

y=3/4,4/5

Hence x ≤ y

Question 9 of 10

9. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

I. 3x2 – 8x + 4 = 0

II. 4y2 – 15y + 9 = 0

if x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or no relation can be established between x and y.

Correct

(e)

3x2 – 8x + 4 = 0

or 3x2 – 6x – 2x + 4 = 0

or (3x – 2) (x – 2) = 0

x=2,2/3

II. 4y2 – 15y + 9 = 0

or 4y2 – 12y – 3y + 9 = 0

or 4y(y – 3) – 3(y – 3) = 0

or (4y – 3) (y – 3) = 0

y = 3/4, 3

Relation cannot be established between x and y.

Incorrect

(e)

3x2 – 8x + 4 = 0

or 3x2 – 6x – 2x + 4 = 0

or (3x – 2) (x – 2) = 0

x=2,2/3

II. 4y2 – 15y + 9 = 0

or 4y2 – 12y – 3y + 9 = 0

or 4y(y – 3) – 3(y – 3) = 0

or (4y – 3) (y – 3) = 0

y = 3/4, 3

Relation cannot be established between x and y.

Question 10 of 10

10. Question

1 points

Category: Aptitude

D.1-5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

All the daily quizzes will be updated in the table in the above link, please do attend the quizzes in a daily manner so that, the chances of scoring in the preliminary examination of the IBPS Clerk 2017 is more. At the end, the ultimate aim is to score more and this may be a start to that achievement in the end.